Solution

Just gotten 2 sol.s from 2 person:

-Kenneth Ong
-Kok Han

Sorry kenneth forgot to save yr sol. and hence unable to post your sol here

And here is the sol. for that qn...


Let x be the number of chocolate
Let y be the number of sweets

Jimmy >> 0.5x + (0.5y - 12)
Ken >> (0.5x - 18) + 0.5y

Jimmy's ratio of sweet : chocolate >> 1 : 7
x _______
y _

that means 7 times of (0.5y-12) equals to 0.5x

Ken's ratio of sweet : chocolate >> 1 : 4
x ____
y _

that means 4 times of 0.5y equals to (0.5x - 18)

0.5x = 7(0.5y - 12)
0.5x = 3.5y - 84
3.5y - 84 = 0.5x
3.5y - 0.5x = 84 >> equation 1

0.5x -18 = 4(0.5y)
0.5x - 18 = 2y
2y = 0.5x - 18
2y - 0.5x = -18 >> equation 2

from equation 2 >> everything multiple by -1 >> 0.5x - 2y = 18 >> equation 3

elimination method

3.5y - 0.5x + 0.5x - 2y = 84 + 18
1.5y = 102
y = 68

since y = number of sweets, total sweets bought is 68

And his sister one...

1. There are more chocs than sweets. Ken ate chocs...sweets to chocs is still 1:4.
Let no. of sweets & chocs Ken/Jim has be 'S' & 'C'. Let diff. between chocs & sweets be 'D'. So...

S - C = D

2a. Jim ate 12 sweets...difference is 'D+12', and this is 6x the final no. of sweets (S-12).

2b. Ken ate 18 chocs...difference has decreased to 'D-18', and this is 3x the final no. of sweets=initial no. of sweets .


2a. D+12 = 6(S-12)
2b. D-18 = 3S

This is a less complicating way of figuring out. Instead of thinking of simultaneous eqns, solving for 'S' is enough by combining the equations together.

Solving, S = 34. multiply that by 2 to get 68 which is the answer.

And another one from a website

http://forums.hardwarezone.com.sg/showpost.php?p=40880768&postcount=40